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RoseWind [281]
3 years ago
10

A machine

Physics
1 answer:
In-s [12.5K]3 years ago
3 0

Answer:

Power_input = 85.71 [W]

Explanation:

To be able to solve this problem we must first find the work done. Work is defined as the product of force by distance.

W = F*d

where:

W = work [J] (units of Joules)

F = force [N] (units of Newton)

d = distance [m]

We need to bear in mind that the force can be calculated by multiplying the mass by the gravity acceleration.

Now replacing:

W = (80*10)*3\\W = 2400 [J]

Power is defined as the work done over a certain time. In this way by means of the following formula, we can calculate the required power.

P=\frac{W}{t}

where:

P = power [W] (units of watts)

W = work [J]

t = time = 40 [s]

P = 2400/40\\P = 60 [W]

The calculated power is the required power. Now as we have the efficiency of the machine, we can calculate the power that is introduced, to be able to do that work.

Effic=0.7\\Effic=P_{required}/P_{introduced}\\P_{introduced}=60/0.7\\P_{introduced}=85.71[W]

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g Two cars, car 1 and car 2 are traveling in opposite directions, car 1 with a magnitude of velocity v1=13.0 m/s and car 2 v2= 7
bogdanovich [222]

Answer:

When they are approaching each other

    f_a = 2228.7 \  Hz

When they are passing  each other

    f_a = 2100Hz

 When they are retreating  from each other

     f_a =  1980.7 Hz

Explanation:

From the question we are told that

     The velocity of car one is  v_1 = 13.0 m/s

      The velocity of car two is  v_2 = 7.22 m/s

     The frequency of sound from car one is  f_e = 2.10 kHz

Generally the speed of sound at normal temperature is  v = 343 m/s

  Now as the cars move relative to each other doppler effect is created and this  can be represented  mathematically  as

              f_a = f_o [\frac{v \pm v_o}{v \pm v_s} ]

Where v_s is the velocity of the source of sound

            v_o is the velocity of the observer of the sound

            f_o is the actual frequence

             f_a  is the apparent frequency

Considering the case when they are approaching each other

        f_a = f_o [\frac{v +  v_o}{v -  v_s} ]

          v_o = v_2  

         v_s = v_1

         f_o = f_e

Substituting value

            f_a = 2100  [\frac{343 +  7.22}{ 343  -  13} ]

              f_a = 2228.7 \  Hz

Considering the case when they are passing  each other    

At that instant

                  v_o = v_s = 0m/s

                   f_o = f_e

               f_a = f_o [\frac{v }{v } ]

              f_a = f_o

Substituting value

             f_a = 2100Hz

Considering the case when they are retreating  from each other    

                f_a = f_o [\frac{v -  v_o}{v +   v_s} ]

          v_o = v_2  

         v_s = v_1

         f_o = f_e      

Substituting value

         f_a = 2100  [\frac{343 -  7.22}{343 +   13} ]    

          f_a =  1980.7 Hz    

7 0
3 years ago
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Explanation:

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=78,394.5 × 2,079.36

=163,010,387.52 kg m/s

This should be your answer.

7 0
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